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Term Life Insurance
CHAPTER IV
FINITE DIFFERENCES. INTERPOLATION
20    FINITE DIFFERENCES
is then completely determined and the value corresponding to any value of x can be obtained.
In the majority of cases, however, this is not the most simple method of working, for other devices can be adopted which will materially shorten the arithmetical work. It is important to note, however, that alternative formulas, in which the same values of the function are used, lead to identical results.
In some cases there is scope for the exercise of the ingenuity of the solver, but usually the problems fall into the main categories which are illustrated in the following examples.


5. Example 1. When n equidistant values of a function are given and it is required to find the value of some intermediate term or terms.
This can be done readily by the application of formula (1) of Chapter III, or by the simpler formula (3). From the given values the successive orders of differences are calculated, and the result is obtained by direct substitution.
Thus, taking the numbers living by the HM table at ages 45, 50, 55, 60 and 65, it is required to find the value for age 57.
In conformance with formula (3) the given values can be denoted by f (0), f (1), ..., so that the required value is f (2.4). Then
    1 f(2.4)=f(0)+2.40f(0)+ 24 x2 4 A2f(0)


2.4x1.4x•4    2.4x1.4x x—•6
+    -

x .f lx) f (x) A'f(x) 03 f lx) f (x)
0 77918 -5123 -1106 -389 +75
1 72 795 - 6229 -1495 - 314  
2 66566 - 7724 -1809    
3 58842 - 9533      
4 49309        

From above:
f(2.4)=77918+2.4(—5123)+1.68(—1106) +'224(—389)—•0336(+75) =77918—12295.2—18581—87.1—2.5
= 63675.1.
The value given by the table is 63677.

6    03 f(0) +    24    A4f (0)•
The working is as follows:

INTERPOLATION    21
The difference between the interpolated value and the true value is due to the fact that the interpolation curve, which is based on the assumption that all differences of higher order than the fourth vanish, represents only approximately the true function.
6. Example 2. When the values given and the value sought constitute a series of equidistant terms.
If there are n terms given of which n - 1 are known, then, as explained in § 3, it must be assumed that the (n - 1)th order of differences is zero.
Thus, using formula (2) of Chapter III, we have
An-if (0)=0=f('n-1)-(n-1)f(n-2)+ (n21)f(n-3)-...
+ (-1)n-l.f (0)
In this equation there is only one unknown quantity and its value can, therefore, be readily obtained. For example, if
f (0) = log 3'50 = '54407,
22    FINITE DIFFERENCES
8. Example 4. If several equidistant values are given, together with one isolated term.
For instance, if three values f (0), f (1) and f (2) are given, together with a further value f (h). Having four values of the function it must be assumed that the fourth order of differences is zero and it remains to find the values of the other three leading differences. The first two leading differences are obtained at once by differencing the first three terms of the series, and the value of the third difis then given by the equation
.f(h)=f (0)+hOf(0)+(2) 02 f(0)+(3) z f(0)
For example, taking the numbers living by the H`' table at ages 45, 46, 47 and 50, it is required to find values for ages 48 and 49.
  0
A2
f(0)=77918 -954
-32
f(1)=76964 -986  
f (2) = 75978    
f (5) = 72795    

f(5)=f(0)+5Af(0)+1060f(0)+10.~3f(0),
f(0)° f (5)—[f(0)+50f(0)+1002f(0)] 10
72795 — [77918 — 4770 — 320]
10
=--- - 3.3.

The table is then completed by addition. Thus:
Age
 f (x) A 4,2
A3
45
 '77918 — 954 — 32 — 3.3
46
 76964 — 986 — 35.3 — 3.3
47
 75978 -1021 3 — 38.6 — 3.3
48
 74956.7 -1059'9 — 41'9  
49
 73896.8 -1101.8    
50
 72795.0      

The work is checked by the reproduction of the value for age 50. The tabular values for ages 48 and 49 are 74957 and 73896 respectively.

    INTERPOLATION    23
'Values precisely the same as those obtained above would have been given if the two missing terms had been inserted by the method described in Example 3. It is instructive to confirm this by actual calculation and to compare the two methods of procedure.
9. Example 5. Subdivision of Intervals.
This problem arises when a series of equidistant terms of a series is given (usually every fifth term or every tenth term) and it is desired to find by interpolation the values of all the intermediate terms.
The simplest method of procedure is to calculate from the given values the differences corresponding to the individual terms of the series (the subdivided differences) and thence to construct the table by summation. The calculation is checked by the reproduction of the values of the original terms.
Thus assume that the given terms are f (0), f (1), ... f (5) and it is desired to complete the series f (0), f (*), f (a), etc. It is conto adopt the notation
f (1) — f (0) = Af (0),
and    f (3) — f (0) = s f (0).
The problem then becomes to express 3f (0), 82f (0), ... in terms of A f (0), Q2 f (0), ....
Writing f (1), f (2), ... in terms of the subdivided differences,
f (0) = f (0)
    f (1)= f (0) + 58f (0) + 1082f (0) + 1083f (0) +    584f (0) +    83/(0)
 24    FINITE DIFFERENCES
Whence the values of 8f (0), 82f (0), ... 8y(0) can readily be obtained.
10. Alternatively the formulas for 8, 82, ... can easily be written down by using the method of Separation of Symbols.
Thus
(1
 + 8)5 f (x) = (1 + 0) f (x).
Therefore
(1
 + 8) f (x) = (1 + A) * f (x),
and  
6 f (x) = [(l +    — 1] f (x)
   
_ [' 2 A — '08 02 + '048 As ... ] f (x).
Hence  
82 f (x) = ['2 A — '08 02 + '048 As ... ]2 f (x)

_ ['0402—.03203+'0256A4...] f(x) and so on.
For convenience the coefficients of 0, As, ... occurring in the values of 8, 82, ... are given, for the intervals 5 and 10, in the foltables.
Subdivision into 5 intervals
  Coefficient of
Value of
  
 
A
 A2 D2 A4 As
a
+ '2
 — '08 + •048 — '0336 +'025536
S2  
+'04
 — •032 +'0256 —'02112
S3    
+'008
 —'0096 +'00960
84      
+ •0016
 — •00256
as         + '00032

Subdivision into 10 intervals
  Coefficient of
Value of
  
 
A
e2
A2
A4
As
d
+'1
 — •045 + •0285 — •0206625 + •01611675
d2   + '01 --•009 +.007725 — 0066975
S3     +'001 — •00135 +'0014625
S4       + '0001 — 00018
as         + 00001



INTERPOLATION    25
11. The following example gives an illustration of the method of working.
Given the present values, at 3 per cent. interest, of an annuity of 1 per annum for 20, 25, ... 45 years, it is required to find the intervalues.
We have
x
 f (x) 0 0' Da D4 05
20
 14'8775 2.5356
— '3483
 •0478 — •0065 •0007
25
 17.4131 2.1873
—'3005
 '0413 —'0058  
30
 19.6004 1.8868
— •2592
 •0355    
35
 21.4872 1.6276
— •2237
      
40
 23.1148 1.4039        
45
 24.5187          

Applying the factors given in the former of the above tables, we have
8 f (20) = [•20 — .0802 + •048A' — •0336A4 + .025536W] f (20)
= •2 x 2.5356 +.08 x •3483+'048 x •0478+'0336 x •0065 + •025536 x •0007
= •5375146752.
Similarly
82 = — .015642784,
84 _
 — •000012192,
83 = •000451520,
84 =
 •000000224.

The table is then constructed by addition from these leading differences as shown below.
It is necessary to consider how many decimal places should be retained in the working. This depends, in the first place, on the degree of accuracy desired in the result. Thus if four decimal places are required in f (x), our result must be given to at least one place more. Further, the range of the formula is 25 terms, and in the final term the values of 8, 82, ... 85 are multiplied respectively
by (21) , (2.25), ... (5 ), which are respectively equal to 25, 300,
2300, 12650, 53130. In view of the magnitude of these coefficients, we shall need to retain at least six decimal places more in the value of 84 than are required in the final value of f (x). Since the coof the lower orders of differences are smaller than 53130,

26    FINITE DIFFERENCES
a correspondingly smaller number of decimal places can be retained. In practice, however, there is little to be gained by cutting down, unless only a rough result is required.
The working of the first five terms in the example is shown below. It will be noticed that the accuracy of the work up to this stage is checked by the exact reproduction of the value of f (25). The interpolated values agree exactly with the true values to four places of decimals.
x
 f (x) g2 g3 $4 b5
20
 14.8775
5375147
—'01564278
 •000451520 —'000012192 '000000224
21
 15.415015
5218719
1519126
 439328 11968 224
22
 15.936887
5066806
1475193
 427360 11744 224
23
 16.443568
4919287
1432457
 415616 11520 224
24
 16.935497
4776041
1390895
 404096 11296 224
25
 17.413101
4636951
1350485
 392800 11072 224

12. Example 6. Lagrange's Theorem.
We have now to consider the construction of an interpolation formula which will apply when the given terms of the series are not equidistant.
Let n values of the function be given, namely
f(a),f(b),f(c),...f(n).
Then the function must be assumed to be a parabolic function of x of degree (n — 1). (See § 3.) Assume therefore that the function can be represented as
f(x)= A(x—b)(x—c)... (x — n)
+ B (x—a)(x—c)... (xn) + C(x —a)(x—b)... (x — n) + etc.,
there being n terms in all, each composed of (n — 1) factors multi-plied by a constant, the values of the constants having yet to be determined. It is clear that the right-hand side of the equation is of degree (n -1).
To find the values of the constants we proceed as follows:
    Put    x = a.
    Then    f(a)=A(a—b)(a—c)...(a—n).

INTERPOLATION    27
Therefore    A=    f (a)
(a — b)(a — c)...(a — n)'
Similarly    B = (b — a)(bf o)... (b — n)'
and so on.
Substituting these values of A, B,... in the original equation, we have
f(x)= f (a) (x — b) (x — 0... (x — n) +f (b)(x—a)(x—c)...(x—n)
    
(a — b) (a — c)...(a — n)    (b — a) (b — c)...(b — n)


+...+f(n)(x—a)(x—b)(x—c)    (1).
(n — a) (n — b)(n — c) ...


By an obvious transformation, the formula can be put in a some-what simpler form for calculations, namely
f (x)    f (a)
(x — a) (x — b) ... (x—n) (x—a)(a—b)(a—c) ...(a—n)
f (b)    f (n)
+ (b — a)(x — b)...(b — n) +    + (n — a) (n — b) (n — c)...(x — n)
    (2).


In memorising the formula it should be noted that the denomiare made up of the product of the algebraic differences of the values of the variable, the term (a — a) being replaced by (x — a) and so on.
13. The formula is somewhat laborious to apply, and careful attention to signs is required, but it is convenient to use where only one or two unknown values of the function are required. Since the assumptions underlying it are precisely similar to those previously explained, its use in any particular case will give identical results with those which can be obtained by the use of the ordinary methods of Finite Differences where a sufficiently high order of differences has been taken into account.
To illustrate this point and to provide an example of the use of the formula, we will calculate by Lagrange's formula the value for age 49 in Example 4. In this case we have
f(0)=77918, f(l)=76964, f(2)=75978, f(6)=72795, and it is required to find the value off (4).

28    FINITE DIFFERENCES
We have accordingly

.f (4)    _    .f (0)
(4—0)(4—1)(4—2)(4—5) (4—0)(0—1)(0—2)(0—5)
f (1)    __ _.f(2)
+(1—0)(4—1)(1—2)(1—5)+(2—0)(2—1)(4—2)(2—5)
+    f(5) (5—0)(5—1)(5—2)(4—5)'
or,    — '2I f (4) = — I f (0) + I f (1) — j~f (2) — I f (5)•
Whence we find f (4) = 73896.8, as before.
It should be noted, as a check on the formula, that the sun of the coefficients of the terms on the right-hand side of the equation must equal the coefficient of the term on the left-hand side of the equation.
14. Problems of interpolation between terms at unequal intercan also be dealt with in a simple way by the formulas of Divided Differences (see Chapter VIII).